## Monday, May 21, 2012

### answer to last week's MGRE Math Beast Challenge

The correct answer was indeed (B)!

Here's MGRE's explanation:

This problem gives us an equilateral triangle and a circle, and tells us that the perimeter of the equilateral is 1.25 times the circumference of the circle.

This is license to plug in. Since both an equilateral and a circle are regular figures—that is, all equilaterals are in the same proportion as all other equilaterals, and all circles are in the same proportion as all other circles—we can be certain that we only need to plug in one set of values in order to be sure of the answer. Because we have regular figures and a way to relate them (triangle perimeter = 1.25 × circumference), we will not need to repeatedly try different values as we often do on Quantitative Comparisons.

We could say the radius of the circle is 2, so the circumference is 4π. Then, the perimeter of the equilateral triangle is (1.25)(4π) = 5π . This isn’t ideal, because then we are stuck with π in the calculations for the triangle, where it is unnecessarily awkward.

We could say circumference = 4 and therefore the perimeter of the triangle = (1.25)(4) = 5. But then the side of the equilateral triangle is 5/3, a non-integer, which is inconvenient. We can avoid this by using larger numbers (multiplied by a factor of 3).

So our smartest numbers are circumference = 12 and therefore the perimeter of the triangle = (1.25)(12) = 15, so each side of the triangle is 5. The height of an equilateral triangle is always (√3)/2 the side length. You can always derive this yourself by splitting the equilateral triangle into two 30–60–90 triangles, with known side ratio 1x : (√3)x : 2x. So, the area of the triangle is

(1/2)bh =

(1/2) • (5) • ((5)((√3)/2)) =

(25√3)/4 =

approx. 10.625. (Use the calculator.)

In the circle, circumference = 12 = 2πr, so

r = 12/(2π) = 6/π. The area of the circle is

π(r^2) = π((6/π)^2) = 36/π = approx. 11.46.

(Use the calculator and the approximation 3.14 for π.)